Test Index

Vector Algebra Part 3

Section: Mathematics

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Question : 7 of 72

Marks: +1, -0

Let

\vec{a}=2\hat{i}+\hat{j}-2\hat{k}, \vec{b}=\hat{i}+\hat{j}

\vec{c}=\vec{a}\times \vec{b}

\vec{d}

be a vector such that

|\vec{d}-\vec{a}|=\sqrt{11},|\vec{c}\times \vec{d}|=3

and the angle between

\vec{c}

\vec{d}

\;\frac{\pi}{4}

\vec{a}\cdot \vec{d}

[24 jan 2026 Shift 1]

0
1
3
11

Solution:

Concept:

Use the relationship between cross product magnitude and dot product to find the unknown dot product.

Explanation:

Given

\vec{a}=2\hat{i}+\hat{j}-2\hat{k}

and

\vec{b}=\hat{i}+\hat{j}

.

First, compute

\vec{c}=\vec{a}\times\vec{b}

.

\vec{c}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{vmatrix} = \hat{i}(1\cdot0 - (-2)\cdot1) - \hat{j}(2\cdot0 - (-2)\cdot1) + \hat{k}(2\cdot1 - 1\cdot1)

.

So

\vec{c}=2\hat{i}-2\hat{j}+\hat{k}

.

Magnitude

|\vec{c}| = \sqrt{4+4+1}=3

.

Given

|\vec{c}\times\vec{d}|=3

and angle

\frac{\pi}{4}

between

\vec{c}

and

\vec{d}

.

Use

|\vec{c}\times\vec{d}|=|\vec{c}||\vec{d}|\sin\theta

.

Thus

3 = 3\cdot|\vec{d}|\cdot\frac{1}{\sqrt{2}} \implies |\vec{d}|=\sqrt{2}

.

Given

|\vec{d}-\vec{a}|=\sqrt{11}

. Square both sides:

|\vec{d}-\vec{a}|^2=11

.

Expand:

|\vec{d}|^2+|\vec{a}|^2-2\vec{d}\cdot\vec{a}=11

.

We have

|\vec{a}|^2=4+1+4=9

, so

2+9-2\vec{a}\cdot\vec{d}=11

.

Simplify:

11-2\vec{a}\cdot\vec{d}=11 \implies -2\vec{a}\cdot\vec{d}=0 \implies \vec{a}\cdot\vec{d}=0

.

Answer:

\vec{a}\cdot\vec{d}=0

Therefore, the correct option is A.

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